If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is `(3)/(2)` units, then its eccentricity is:

Options

A. `(1)/(3)`
B. `(2)/(3)`
C. `(1)/(9)`
D. `(1)/(4)`

Answer

To find the eccentricity of the ellipse given the conditions in the problem, we can follow these steps:

### Step 1: Understand the given information
We know:
- The length of the latus rectum \( L = 4 \) units.
- The distance between a focus and its nearest vertex on the major axis \( d = \frac{3}{2} \) units.

### Step 2: Set up the equations
For an ellipse in the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (where \( a > b \)):
- The length of the latus rectum is given by \( L = \frac{2b^2}{a} \).
- The distance from the focus to the vertex is given by \( d = a - ae = a(1 - e) \).

### Step 3: Write the equations based on the given information
From the length of the latus rectum:
\[ \frac{2b^2}{a} = 4 \quad \Rightarrow \quad b^2 = 2a \quad \text{(Equation 1)} \]

From the distance from the focus to the vertex:
\[ a(1 - e) = \frac{3}{2} \quad \Rightarrow \quad a = \frac{3}{2(1 - e)} \quad \text{(Equation 2)} \]

### Step 4: Use the relationship between \( a \), \( b \), and \( e \)
We know that:
\[ e^2 = 1 - \frac{b^2}{a^2} \quad \Rightarrow \quad b^2 = a^2(1 - e^2) \quad \text{(Equation 3)} \]

### Step 5: Substitute Equation 1 into Equation 3
Substituting \( b^2 = 2a \) into Equation 3:
\[ 2a = a^2(1 - e^2) \]
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ 2 = a(1 - e^2) \quad \Rightarrow \quad a = \frac{2}{1 - e^2} \quad \text{(Equation 4)} \]

### Step 6: Substitute Equation 2 into Equation 4
Now, we can equate the two expressions for \( a \):
\[ \frac{3}{2(1 - e)} = \frac{2}{1 - e^2} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ 3(1 - e^2) = 4(1 - e) \]
Expanding both sides:
\[ 3 - 3e^2 = 4 - 4e \]
Rearranging gives:
\[ 3e^2 - 4e + 1 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = -4, c = 1 \):
\[ e = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6} \]
Calculating the two possible values:
1. \( e = \frac{6}{6} = 1 \) (not valid for an ellipse)
2. \( e = \frac{2}{6} = \frac{1}{3} \)

### Final Answer
Thus, the eccentricity \( e \) of the ellipse is:
\[ \boxed{\frac{1}{3}} \]